Suppose $X$ is an irreducible Noetherian normal scheme. Let $D$ be an irreducible closed codimension $1$ subset/subscheme of $X$.
Recall that the irreducible Weil divisor $[D]$ is locally principal if there is an open cover $U_i$ of $X$ such that $[D\cap U_i] = \operatorname{div} f_i$ for some rational function $f_i$ over $U_i$, and $D$ is a locally principal closed subscheme of $X$ if there exists an open cover $U_i$ of $X$ such that $D\cap U_i$ is cut out by one equation.
Question. Does $[D]$ being locally principal imply that $D$ is a locally principal closed subscheme?
Thoughts. I expect that over $U_i$, the ideal of $D\cap U_i$ is given by $(f_i)$. By the assumption, $f_i$ generates the principal maximal ideal in $\mathcal{O}_{X,\eta}$, where $\eta$ denotes the generic point of $D$. (This appears strange to me because it does not depend on the index $i$.) This is only a necessary condition for $f_i$ to cut out $D\cap U_i$ in $U_i$, and I am not sure how to prove that it is sufficient.
Could anyone answer the question and also clarify the relationship between the two notions of "locally principal"? Thanks in advance!
Motivation. I am trying to show that $D:=\overline{\{[(x,z)]\}} \subset X:=\operatorname{Spec} k[x,y,z]/(xy-z^2)$ gives an irreducible divisor that is not locally principal. By looking at the stalk at $[(x,y,z)]$, we see that $(x,z)$ is not principal in the local ring $(k[x,y,z]/(xy-z^2))_{(x,y,z)}$, so $D$ is not a locally principal closed subscheme. But I am not sure whether this also shows that $[D]$ is not a locally principal divisor.